Solving a System of Equations Using a Matrix

Key Questions

  • How do I solve a system of equations using an augmented matrix?

    An augmented matrix contains the coefficients of the unknowns and the "pure" coefficients. You can manipulate the rows of this matrix (elementary row operations) to transform the coefficients and to "read", at the end, the solutions of your system.

    The two row operations allowed are:
    1) swap rows;
    2) take the elements of a row, multiply them by a scalar and sum them to the corresponding element of another row.

    I show you an example. Let us have the following system:
    enter image source here
    You can see that now Row 2 contains a zero. This is good because you eliminate one unknown from the equation represented by Row 2 and so you can "see" the value of the remaining one.
    enter image source here

    You can now apply the same idea to bigger systems with more equations and unknowns.

    Gió · 1 · Dec 30 2014
  • How do I solve a system of equations using inverse matrices?

    Consider a normal equation in #x# such as:
    #3x=6#
    To solve this equation you simply take the #3# in front of #x# and put it, dividing, below the #6# on the right side of the equal sign.
    #x=6/3=3^-1*6=2#
    at this point you can "read" the solution as: #x=2#.
    With a system of #n# equations in #n# unknowns you do basically the same, the only difference is that you have more than 1 unknown (and equation) that can now be represented by matrices and by the inverse matrix in place of the coefficient to the -1 (in our example is #3^-1#).

    You can use matrices and change your system in a matrix equation :
    If you have the following system:
    enter image source here
    (where #a,b,c,d,e,g# are real numbers)
    you can change it in a matrix equation:
    enter image source here
    Where #A# is the matrix of the coefficients of the unknowns, #U# is the column of the unknown and #B# is the column of the pure coefficients (without unknowns).

    You can check that this representation with matrices represents the system by doing the multiplication #A*U# and setting it equal to #B# you'll get back your original system!!!

    Now, to solve your matrix equation #A*U=B# you can multiply both sides by the inverse of #A#, i.e. #A^-1#
    enter image source here
    (Remembering that #I# is the identity matrix .

    For example:
    enter image source here
    So:
    #x=-3#
    #y=5#

    Gió · 1 · Dec 29 2014
  • How do I solve a system of 3 equations with a matrix?

    I would use Cramer's Rule .
    This rule is based upon using the matrix associated with the coefficients of the unknowns and of the pure coefficients of your system, evaluate their determinants and combine them to evaluate the values of the unknowns.

    Consider an example:

    enter image source here

    Now we consider 3 other matrices, #A_x, A_y and A_z# and their determinant. These matrices are obtained by substituting each column of #A# with the column of pure coefficients (the ones without unknown):

    enter image source here

    We evaluate the three determinants for these matrices:

    enter image source here

    Finally we can calculate the values of the unknowns as:

    #x=det(A_x)/(det(A))=(-60)/-60=1#
    #y=det(A_y)/(det(A))=(-240)/-60=4#
    #z=det(A_z)/(det(A))=(120)/-60=-2#

    Your final result is:
    #x=1#
    #y=4#
    #z=-2#

    Gió · 1 · Dec 25 2014
  • How do I solve a system of 2 equations using a matrix?

    You can solve a square system of 2 linear equations using Cramer's Rule or Reduced Row Echelon Form.

    A square system has the same number of equations as variables. A square system can be classified as independent, dependent, or inconsistent.

    If you have more variables than equations, you would have an underdetermined system and will be classified as dependent or inconsistent.

    If you have fewer variables than equations, you would have an overdetermined system and will be classified as independent, dependent, or inconsistent. However in the case of just 2 equations, it cannot be dependent.

    Reduced Row Echelon Form (RREF) is solving the system similar to using the elimination method to solve a system of linear equations.

    Cramer's Rule is solving the system using determinants. Cramer's Rule is not necessarily faster than RREF, but it doesn't require thinking because it is a formula; this is good for calculators and computers.

    Solving with a matrix is usually quicker than solving with variables because there is less to write out. Furthermore, we have other techniques to solve the system over elimination and substitution.

    So search for RREF and Cramer's Rule as there are plenty of examples available.

    Amory W. · · Sep 1 2014

Questions

Matrix Row Operations